RS AGGARWAL CLASS 9 CHAPTER 10 QUADRILATERALS EXERCISE 10A

 EXERCISE 10A

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Q1 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 1:

Three angles of a quadrilateral are 75°, 90° and 75°. Find the measure of the fourth angle.

Answer 1:

Given: Three angles of a quadrilateral are 75°, 90° and 75°.

Let the fourth angle be x.

Using angle sum property of quadrilateral,

⇒75°+90°+75°+x=360°

⇒240°+x=360°

⇒x=360°−240°

⇒x=120°

⇒x=360°-240°

⇒x=120°

So, the measure of the fourth angle is 120°

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Q2 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 2:

The angles of a quadrilateral are in the ratio  2: 4 : 5 : 7. Find the angles.

Answer 2:

Let $\angle$A = 2x^∘​.
Then $\angle$B = (4x)^∘; $\angle$C = (5x)^∘ and $\angle$D = (7x)^∘
Since the sum of the angles of a quadrilateral is 360^o, we have:
2x + 4x + 5x + 7x = 360^∘   
⇒ 18 x = 360^∘ ​
⇒ x = 20^∘
∴  $\angle$A = 40^∘; $\angle$B = 80^∘; $\angle$C = 100^∘; $\angle$D = 140^∘

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Q3 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 3:

In the adjoining figure, ABCD is a trapezium in which AB || DC. If ∠A = 55° and ∠B = 70°, find ∠C and ∠D.

Answer 3:

  We have AB || DC.

 $\angle$ A  and  $\angle$ D are the interior angles on the same side of transversal line AD, whereas $\angle$ B and  $\angle$ C are the interior angles on the same side of transversal line BC. 
Now,  $\angle$A + $\angle$D = 180^∘
⇒  $\angle$D = 180^∘ − $\angle$A  
∴ $\angle$ D = 180^∘ − 55^∘ = 125^∘

Again , $\angle$ B + $\angle$C = 180^∘
​⇒  $\angle$C  = 180^∘ − $\angle$B  
∴ $\angle$ C = 180^∘ −  70^∘ = 110^∘
 

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Q4 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 4:

In the adjoining figure, ABCD is a square and ∆EDC is an equilateral triangle. Prove that (i) AE = BE, (ii) ∠DAE = 15°.

Answer 4:

Given:  ABCD is a square in which AB = BC = CD = DA. ∆EDC is an equilateral triangle in which ED = EC = DC and  
$\angle$ EDC  =   $\angle$ DEC = $\angle$DCE =  60^∘.
To prove:  AE = BE and $\angle$DAE = 15^∘​
Proof: In ∆ADE and ∆BCE, we have:
AD = BC              [Sides of a square]

DE = EC​             [Sides of an equilateral triangle]
$\angle$ADE  = $\angle$BCE = 90^∘ +  60^∘ = 150^∘​ 

∴ ∆ADE ≅  ∆BCE
i.e., AE =  BE  

Now, $\angle$ADE = 150^∘
DA = DC     [Sides of a square]
DC = DE      [Sides of an equilateral triangle]
So, DA = DE
∆ADE and ∆BCE are isosceles triangles.
i.e., ∠DAE = ∠DEA $= \frac{1}{2}\left(180^{\circ} − 150^{\circ} \right) $

$=\frac{ 30^{\circ}}{2} $=15°
 

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Q5 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 5:

In the adjoining figure, BM ⊥ AC and DN ⊥ AC. If BM = DN, prove that AC bisects BD.

Answer 5:

Given: A quadrilateral ABCD, in which BM ​⊥ AC and DN ⊥ AC and BM = DN.
To prove: AC bisects BD; or DO = BO

Proof:
Let AC and BD intersect at O.
Now, in ∆OND and ∆OMB, we have:
∠OND = ∠OMB                 (90^o each)
∠DON = ∠ BOM                  (Vertically opposite angles)
Also, DN = BM                         (Given)
i.e., ∆OND ≅ ∆OMB             (AAS congurence rule)
∴ OD = OB                          (CPCT)
​Hence, AC bisects BD.

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Q6 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 6:

In the given figure, ABCD is a quadrilateral in which AB = AD and BC = DC. Prove that
(i) AC bisects ∠A and ∠C,
(ii) BE = DE,
(iii) ∠ABC = ∠ADC.

Answer 6:

​Given:  ABCD is a quadrilateral in which AB = AD and BC = DC  
(i)
In ∆ABC and ∆ADC, we have:
AB = AD                                                  (Given)

BC = DC                                                 (Given)
AC is common.
i.e., ∆ABC ≅ ∆ADC                                    (SSS congruence rule)

∴ ∠BAC = ∠DAC and ∠BCA = ∠D​CA        (By CPCT)
Thus, AC bisects ∠A and ∠ C.

(ii)
Now, in ∆ABE and ∆ADE, we have:
  AB = AD                                      (Given)​
∠BAE = ∠DAE​                               (Proven above)
 AE is common.
∴ ∆ABE ≅  ∆ADE                          (SAS congruence rule)
⇒ BE = DE                                                 (By CPCT)
 
(iii)   ∆ABC ≅  ∆ADC                  (Proven above)
∴ ∠ABC = ∠AD​C                           (By CPCT)​
 

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Q7 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 7:

In the given figure, ABCD is a square and ∠PQR = 90°. If PB = QC = DR, prove that
(i) QB = RC,
(ii) PQ = QR,
(iii) QPR = 45°.

Answer 7:

Given: ABCD is a square and ∠PQR = 90°.
Also, PB = QC = DR

(i) We have:
  BC = CD                (Sides of square)
  CQ = DR                   (Given)
  BC = BQ + CQ
⇒ CQ = BC − BQ
∴ DR = BC − BQ           ...(i)
   
Also, CD = RC+ DR
∴ DR = CD −  RC = BC − RC            ...(ii)
From (i) and (ii), we have:
BC − BQ = ​BC − RC
∴ BQ = RC

(ii) In ∆RCQ and ∆QBP, we have:
PB = QC   (Given)
BQ = RC  (Proven above)
∠RCQ = ∠QBP   (90^o each)
i.e., ∆RCQ ≅ ∆QBP       (SAS congruence rule)
∴ QR =  PQ                (By CPCT)

(iii) ∆RCQ ≅ ∆QBP and QR = PQ (Proven above)
∴ In ∆RPQ, ∠QPR = ∠QRP = $\frac{1}{2}\left(180° - 90°\right) = \frac{90°}{2} = {45}^{°}$

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Q8 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 8:

If O is a point within a quadrilateral ABCD, show that OA + OB + OC + OD > AC + BD.

Answer 8:

Let ABCD be a quadrilateral whose diagonals are AC and BD and O is any point within the quadrilateral. 
Join O with A, B, C, and D.
We know that the sum of any two sides of a triangle is greater than the third side.
So, in ∆AOC, OA + OC > AC

Also, in ∆ BOD, OB + OD > BD

Adding these inequalities, we get:

(OA + OC) + (OB + OD) > (AC + BD)
⇒ OA + OB + OC + OD > AC + BD


 

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Q9 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 9:

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that:
(i) AB + BC + CD + DA > 2AC
(ii) AB + BC + CD > DA
(iii) AB + BC + CD + DA > AC + BD

Answer 9:

Given: ABCD is a quadrilateral and AC is one of its diagonal.

(i)  We know that the sum of any two sides of a triangle is greater than the third side.
In ∆ABC, AB + BC > AC            ...(1)

In ∆ACD, CD + DA > AC            ...(2)

​Adding inequalities (1) and (2), we get:

AB + BC + CD + DA > 2AC 

(ii) In ∆ABC, we have :
 ​  AB + BC > AC            ...(1)
  We also know that the length of each side of a triangle is greater than the positive difference of the length of the other two sides.
   In ∆ACD, we have:​
  AC > |DA − CD|​        ...(2)
   From (1) and (2), we have:
   AB + BC > |DA − CD|​
  ⇒ AB + BC + CD > DA

(iii) In ∆ABC, AB + BC > AC

    In ∆ACD, CD + DA > AC

    In ∆ BCD, BC + CD > BD

   In ∆ ABD, DA + AB > BD
   ​Adding these inequalities, we get:
   2(AB + BC + CD + DA) > 2(AC + BD)
   ⇒ (AB + BC + CD + DA) > (AC + BD)
 

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Q10 | Ex-10A | RS AGGARWAL | Class 9 | QUADRILATERALS | Chapter 10 | myhelper

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Question 10:

Prove that the sum of all the angles of a quadrilateral is 360°.

Answer 10:

Let ABCD be a quadrilateral and ∠1, ∠2, ∠3 and ∠4  are its four angles as shown in the figure.
Join BD which divides ABCD in two triangles, ∆ABD and ∆BCD.
In ∆ABD, we have:
∠1 + ∠2 + ​∠A = 180^o      ...(i)
In ​∆BCD, we have:
∠3 + ∠4 + ∠C = 180^o     ...(ii)
On adding (i) and (ii), we get:

(∠1 + ∠3) + ∠A + ∠C + (∠4 + ∠2) ​= 360^o  
⇒ ∠A + ∠C  + ∠B + ∠D = 360^o                         [ ∵ ∠1 + ∠3 = ∠B; ∠4 + ∠2 = ∠D]
∴ ∠A + ∠C  + ∠B + ∠D = 360^o